\(\frac{x^4+6x^3+9x^2-1}{x^4+6x^3+7x^2-6x+1}=\frac{\left(x^2\right)^2+2.x^2.3x+\left(3x\right)^2-1}{\left(x^2\right)^2+2.x^2.3x+\left(3x\right)^2-2x^2-6x+1}\)

\(=\frac{\left(x^2+3x\right)^2-1}{\left(x^2+3x\right)^2-2\left(x^2+3x\right)+1}\)

\(=\frac{\left(x^2+3x-1\right)\left(x^2+3x+1\right)}{\left(x^2+3x-1\right)^2}=\frac{x^2+3x+1}{x^2+3x-1}\)

\(B=x^2-3x+xy-3y\)

\(B=x.\left(x-3\right)+y.\left(x-3\right)=\left(x+y\right).\left(x-3\right)\)

\(D=x^3-3x^2+3x+100=x^3-3x^2+3x-1+101=\left(x-1\right)^3+101=100^3+101=1000101\)

\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\)

Đặt \(x^2+5x+4=t\)

\(\Rightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\)

\(=t.\left(t+2\right)-15\)

\(=t^2+2t+1-16\)

\(=\left(t+1\right)^2-4^2\)

\(=\left(t-3\right)\left(t+5\right)\)

\(=\left(x^2+5x+1\right)\left(x^2+5x+9\right)\)

Ta có :

\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\)

\(=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-15\)

\(=\left[x\left(x+4\right)+1\left(x+4\right)\right]\left[x\left(x+3\right)+2\left(x+3\right)\right]-15\)

\(=\left(x^2+4x+x+4\right)\left(x^2+3x+2x+6\right)-15\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\)

\(=\left(x^2+5x+4\right)\left[\left(x^2+5x+4\right)+2\right]-15\)(1)

Đặt \(x^2+5x+4=y\)thì (1) trở thành :

\(y\left(y+2\right)-15\)

\(=y^2+2y-15\)

\(=y^2+5y-3y-15\)

\(=\left(y^2+5y\right)-\left(3y+15\right)\)

\(=y\left(y+5\right)-3\left(y+5\right)\)

\(=\left(y-3\right)\left(y+5\right)\)(2)

Thay \(y=x^2+5x+4\)thì (2) trở thành:

\(\left(x^2+5x+4-3\right)\left(x^2+5x+4+5\right)\)

\(=\left(x^2+5x+1\right)\left(x^2+5x+9\right)\)

\(\frac{x^3+x^2-4x-4}{x^3+8x^2+17x+10}\)

\(=\frac{x^2\left(x+1\right)-4\left(x+1\right)}{x^3+x^2+7x^2+7x+10x+10}\)

\(=\frac{\left(x^2-4\right)\left(x+1\right)}{\left(x+1\right)\left(x^2+7x+10\right)}\)

\(=\frac{x^2-4}{x^2+7x+10}\)

\(=\frac{x^2-4}{x^2+5x+2x+10}\)

\(=\frac{\left(x-2\right)\left(x+2\right)}{x\left(x+5\right)+2\left(x+5\right)}\)

\(=\frac{x-2}{x+5}\)

tui 2k5 nè , kb ko ?

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ko đnagư câu hỏi linh tinh